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Math Help - Geometric Progression

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    Geometric Progression

    A geometric progression series has first term 2 with common ratio as -1/2(x + 1). When x becomes 1/3, find the sum of all odd-numbered terms of the series. Answer given as 18/5 - not sure that it is correct! Thanks
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    Quote Originally Posted by puggie View Post
    A geometric progression series has first term 2 with common ratio as -1/2(x + 1). When x becomes 1/3, find the sum of all odd-numbered terms of the series. Answer given as 18/5 - not sure that it is correct! Thanks
    \sum_{k=0}^{\infty}\left(\frac{-(x+1)}{2}\right)^k
    =1-\frac{x+1}{2}+\frac{(x+1)^2}{4}-\frac{(x+1)^3}{8}+...

    So the odd numbered terms are 1+\frac{(x+1)^2}{4}+\frac{(x+1)^4}{16}+... which are the first, third, and 5th terms and so on

    So this is just a geometric series with the ratio = \frac{(x+1)^2}{4} so when x=\frac{1}{3}

    The ratio is \frac{(\frac{1}{3}+1)^2}{4}=\frac{\frac{16}{9}}{4}  =\frac{4}{9}

    So the sum of a geometric series with ratio = \frac{4}{9} is \frac{1}{1-\frac{4}{9}}=\frac{9}{5}

    Sorry I didn't see that the first term was 2, being as this is the case the answer must be doubled and we get the correct answer
    Last edited by artvandalay11; May 22nd 2009 at 04:49 AM. Reason: misread the question
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by puggie View Post
    A geometric progression series has first term 2 with common ratio as -1/2(x + 1). When x becomes 1/3, find the sum of all odd-numbered terms of the series. Answer given as 18/5 - not sure that it is correct! Thanks
    plug in x = 1/3, you find that r = -2/3

    now, we want all odd terms, first note that these are the positive terms, secondly, note that the common ratio between consecutive odd terms is r^2. i hope you can see why. hence, the odd terms form a geometric series with first term a = 2 and common ratio r = 4/9. you should be able to do the rest. we get 18/5 as desired
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